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Viewing Single Post From: The giant space ship example  

Chris HoStuart  Dec 7 2011, 08:43 PM 
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I haven't said anything at all about how long it would take. But I'll do so now, just quickly, for the first time. To answer this, we need to some assumptions.
From post #5 in the thread, mass of the atmosphere is 5.4e17 kg. Heat capacity C_{p} is 1 J/g/K, or 1000 J/kg/K. So the total atmosphere heat capacity is Energy being lost to space is 6.58e15 W, so the cooling rate is [indentblock]I've fixed up an error in my calculations, where I had the heat capacity ten times too high[/indentblock] However, this calculation has ignored the inversion layer that would form at the bottom of the atmosphere. In practice it would drop in temperature faster than this until you get the inversion in approximate equilibrium with a slow diffusion of heat from higher altitudes. Calculating that is just more distraction from the original problem posed.
Yes indeed. Nighttime inversions over the desert can be very shallow, just a few meters even; and that cools much more quickly than the entire atmosphere. A solar eclipse would get a rapid inversion forming also. (I've been in a total solar eclipse, but it was heavily overcast at the time, unfortunately. It went dark like at night, but there was not much temperature drop. Nor would I expect much drop in those conditions of heavy cloud.) Just for the sake of argument, let's assume that about the bottom 4 meters of atmospheres is the effective heat capacity as temperatures start to fall. At pressure 0.3 atm, that's about 1.2 kg of air per square meter, so the total effective heat capacity over the starship surface is roughly 2e17 J/K. Using this value in the calculation above give temperature falling initially at 1 degree every 300 seconds or so; or 5 minutes for a degree. That's a really rough ball park calculation. If you have a poor thermal contact with the surface, then ground temperature will drop much faster, but what we feel is the lowest levels of air, and that would drop more slowly.
There's no atmosphere there; the only heat sink is the top few centimeters of regolith. I think it falls much faster than that even, but I have no figures.
This is the first post in which I have said how long cooling takes.
The power requirements to maintain a given temperature depend crucially on the surface temperature, the amount of atmosphere, and the atmospheric composition. The rate of cooling when you switch off the generators depends on the same things, though the initial cooling rate will also depend on thermal contact between surface and atmosphere. Any additional surface heat capacity also matters, but it would most likely be negligible. Cheers  Chris Edited by Chris HoStuart, Dec 7 2011, 09:02 PM.

The giant space ship example · Physical theory for climate 